Factoring Polynomials How To Factor 2x⁴ + 11x² + 5

Hey guys! Today, we're diving into the fascinating world of polynomials, specifically how to factor them completely. We'll break down a common problem and provide a step-by-step guide to help you master this essential math skill. Let's tackle this problem together and make sure you understand each step clearly. Factoring polynomials might seem daunting at first, but with the right approach, it can become a breeze. So, let's jump right in and get started!

Understanding the Problem

So, the big question we're tackling today is: What is the completely factored form of the polynomial $2x^4 + 11x^2 + 5$? We have four options to choose from:

A. $(2x + 1)(x + 5)$ B. $(2x^2 + 1)(x^2 + 5)$ C. $(2x + 5)(x + 1)$ D. $(2x^2 + 5)(x^2 + 1)$

Before we dive into solving this, let's understand what factoring a polynomial actually means. Factoring is like reverse multiplication. Think of it as breaking down a number or an expression into smaller parts (factors) that, when multiplied together, give you the original number or expression. For example, factoring the number 12 gives you 3 and 4 because 3 * 4 = 12. Similarly, we're going to break down this polynomial into its constituent factors.

Now, looking at our polynomial $2x^4 + 11x^2 + 5$, you might notice it's not a typical quadratic equation, but it has a structure that we can work with. Specifically, it's a quadratic in disguise! Notice the powers of x: we have $x^4$, $x^2$, and a constant term. This is similar to the form $ax^2 + bx + c$, which we are familiar with. To make it even clearer, we can use a substitution to transform it into a more recognizable form. This is a common trick in algebra, and it's super useful for making complex problems simpler. By understanding the structure of the polynomial and using clever substitutions, we can make the factoring process much more manageable. So, keep this in mind as we move forward – spotting patterns and using substitutions can be your best friends when it comes to tackling tough math problems.

Step 1: Substitution for Simplification

Okay, so we've established that our polynomial $2x^4 + 11x^2 + 5$ is a bit like a quadratic equation in disguise. To make this clearer and easier to work with, we're going to use a nifty trick called substitution. This involves replacing a part of the expression with a single variable, which simplifies the whole thing and makes it look more familiar. In this case, the part we're going to substitute is $x^2$.

Let's say we let $y = x^2$. Now, think about what happens when we square y: $y^2 = (x^2)^2 = x^4$. See how that neatly matches the first term in our polynomial? This is exactly what we want!

Now, we can rewrite our original polynomial $2x^4 + 11x^2 + 5$ using this substitution. Instead of $x^4$, we'll write $y^2$, and instead of $x^2$, we'll write y. So, the polynomial becomes:

$2y^2 + 11y + 5$

Wow, doesn't that look much more manageable? This new form is a standard quadratic equation, which we're probably pretty comfortable with factoring. By using this simple substitution, we've transformed a more complex-looking polynomial into something much more familiar and approachable. This is a powerful technique in algebra because it allows us to apply the methods we already know to solve problems that initially seem intimidating. The key takeaway here is that substitution can simplify things dramatically, making the problem-solving process much smoother.

Step 2: Factoring the Quadratic

Alright, we've successfully transformed our original polynomial into a more manageable quadratic equation: $2y^2 + 11y + 5$. Now, the next step is to factor this quadratic. There are a few different methods you can use to factor quadratics, but we'll go through a common and effective one – the factoring by grouping method.

The goal here is to break down the middle term (11y) into two terms that allow us to factor by grouping. To do this, we need to find two numbers that:

1. Multiply to give the product of the leading coefficient (2) and the constant term (5), which is 2 * 5 = 10.
2. Add up to the middle coefficient, which is 11.

Think about it for a moment. Which two numbers fit this criteria? The numbers 10 and 1 work perfectly! 10 * 1 = 10 and 10 + 1 = 11. Great!

Now, we can rewrite the middle term 11y as 10y + y. So, our quadratic equation becomes:

$2y^2 + 10y + y + 5$

Next, we factor by grouping. We look at the first two terms and the last two terms separately:

• From the first two terms, $2y^2 + 10y$, we can factor out a 2y, giving us $2y(y + 5)$.
• From the last two terms, $y + 5$, we can factor out a 1 (which doesn't change anything but helps us see the pattern), giving us $1(y + 5)$.

Now, we have:

$2y(y + 5) + 1(y + 5)$

Notice that both terms now have a common factor of $(y + 5)$. We can factor this out, which gives us:

$(2y + 1)(y + 5)$

And there you have it! We've successfully factored the quadratic equation $2y^2 + 11y + 5$ into $(2y + 1)(y + 5)$. Remember, the key to this method is finding the right numbers to split the middle term and then using grouping to simplify. Now, we're not quite done yet, because we need to go back to our original variable, x. But we're definitely on the right track!

Step 3: Substituting Back

Okay, we've made some great progress! We factored the quadratic equation in terms of y, and we've got $(2y + 1)(y + 5)$. But remember, our original problem was in terms of x, so we need to substitute back to get our final answer. This is a crucial step, so let's make sure we do it correctly.

We initially made the substitution $y = x^2$. So, now we're going to reverse that and replace every y in our factored expression with $x^2$. This means:

• Replace the y in $(2y + 1)$ with $x^2$, giving us $(2x^2 + 1)$.
• Replace the y in $(y + 5)$ with $x^2$, giving us $(x^2 + 5)$.

Putting these together, our factored expression becomes:

$(2x^2 + 1)(x^2 + 5)$

And there we have it! We've successfully substituted back and now have our polynomial factored in terms of x. This step is super important because it brings us back to the original problem and gives us the answer in the correct form. It's easy to forget this step, especially when you're in the middle of a long problem, so always double-check that you've substituted back at the end. Now that we've got our factored form, let's compare it to the answer choices and see which one matches.

Step 4: Identifying the Correct Option

Alright, we've done the hard work and factored our polynomial! We found that the completely factored form of $2x^4 + 11x^2 + 5$ is $(2x^2 + 1)(x^2 + 5)$. Now, let's take a look at our answer choices and identify the correct option:

A. $(2x + 1)(x + 5)$ B. $(2x^2 + 1)(x^2 + 5)$ C. $(2x + 5)(x + 1)$ D. $(2x^2 + 5)(x^2 + 1)$

Comparing our factored form $(2x^2 + 1)(x^2 + 5)$ with the options, it's clear that:

• Option B, $(2x^2 + 1)(x^2 + 5)$, matches perfectly!

So, the correct answer is B. Awesome!

It's always a good idea to double-check your work, especially in math. Make sure each step makes sense, and if you have time, you can even multiply the factors back together to see if you get the original polynomial. This is a great way to ensure you haven't made any mistakes along the way. In this case, if we were to multiply $(2x^2 + 1)(x^2 + 5)$, we would indeed get back to $2x^4 + 11x^2 + 5$, confirming that our factoring is correct. Identifying the correct option is the final step in the problem-solving process, and it's super satisfying when you see your hard work pay off! So, give yourself a pat on the back – you've just successfully factored a polynomial!

Conclusion

So, to wrap things up, we've successfully factored the polynomial $2x^4 + 11x^2 + 5$. Remember, the correct answer is B. $(2x^2 + 1)(x^2 + 5)$. We tackled this problem by using a combination of clever techniques, including substitution and factoring by grouping. Let's quickly recap the steps we took:

1. Substitution for Simplification: We substituted $y = x^2$ to transform the polynomial into a more manageable quadratic form.
2. Factoring the Quadratic: We factored the quadratic $2y^2 + 11y + 5$ into $(2y + 1)(y + 5)$ using the factoring by grouping method.
3. Substituting Back: We substituted back $x^2$ for y, giving us $(2x^2 + 1)(x^2 + 5)$.
4. Identifying the Correct Option: We matched our factored form with the answer choices and found that option B was the correct one.

Factoring polynomials is a fundamental skill in algebra, and it's something that comes up time and time again in higher-level math courses. By understanding the techniques and practicing regularly, you'll become more confident and proficient at it. Remember, problems that seem complex at first can often be simplified with the right approach. Substitution is a powerful tool, and mastering factoring by grouping can help you tackle a wide range of quadratic equations. So, keep practicing, and you'll be factoring polynomials like a pro in no time! And hey, if you ever get stuck, don't hesitate to review these steps or ask for help. We're all in this together, and with a little effort, you can conquer any math challenge!